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GR9677 #61
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Electromagnetism$\Rightarrow$}Gauss Law

Recall Gauss Law $\vec{E} \cdot d\vec{A} = q/\epsilon_0$. Thus, $E(4\pi r^2)=\int_0^{R/2}A r^2 4\pi r^2 dr = 4\pi A (R/2)^5/5$. Solving for E, one has $E=A/5(R/2)^3$, as in choice (B).

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rabia
2012-10-11 01:04:05
I have to solve a problem by using Coulumn' law given rho=rho not (1-r/R) and find Q from given expression.
physicsphysics
2011-10-11 07:22:32
Does anybody have a problem by using Coulumn' law?
I used $\vec{E}=\frac{1}{4\Pi\e0}\int_0^\R/2\frac{Ar^2}{r^2}r^2Sin\Theta dr d\Theta d\Phi$. The answer is $\frac{AT^4}{24e0}$. Can anybody explain what I am wrong?
 3Danyon2011-10-13 08:44:10 I made the same mistake initially. The thing is, one has to calculate the charge first and then divide by the $(R/2)^2$. Otherwise you change the integrand and get the wrong numerical factor. To spell it out, the field is given by: $\vec{E}=\frac{1}{4\pi \epsilon_0} \frac{Q}{(R/2)^2} = \frac{1}{4\pi \epsilon_0 (R/2)^2} \cdot \int (A r^2) \cdot r^2 \sin{\theta} dr d\theta d\phi$
phoxdie
2010-11-10 15:20:06
Sorry to get hung up on this, but it was very confusing the first time I looked at this solution. There is a small "typo" in the solution. A factor of epsilon is dropped for no reason.
Kabuto Yakushi
2010-09-19 10:28:26
Another method for solving this problem would be to take the volume integral of $\rho$ to solve for Q. Then to plug Q into Gauss' law.

$Q=\int_0^{2\pi}\int_0^\pi\int_0^{\frac{R}{2}}{\rho}d\tau$

where $d\tau=r^2sin(\theta)drd{\theta}d{\phi}$

plug Q into Gauss' law:

$E=\frac{Q}{{\pi}r^2{\epsilon_0}}$

solving one gets choice B).

Yosun's solution is faster, but if using the shell method for integration doesn't come to mind, one must resort to the long method.
tsharky87
2009-11-06 08:46:03
I was confused about the fact that I seemed to be getting a $1/160$, so here's an explanation for those of you who were as confused as azot and I:

I was getting my r's confused in Yosun's solution, I just took $r=R$ and canceled away. The reason this is wrong is that the radius that you are taking the area with when multiplying by $\vec{E}$ is actually $R/2$. So if you set $r=R/2$ you should arrive at the solution.
jw111
2008-11-05 15:43:14
in addition, if you set A=1, R=1, epsilon=1

it will be easier

$E*4pi*\frac{1}{2^2}=\int r^2*4pi*r^2 dr$

then
$E \frac{1}{2^2} = \frac{1}{5} \frac{1}{2^5}$

E= 1/40
 pam d2011-09-17 16:24:27 nice
gliese876d
2008-10-11 20:56:52
hmm. I'm just wondering... I had the following formula memorized for similar situations:

E=$\frac{Qr}{4piepsilonR^3}$

Where little r is the radius of the point from the center and R is the total radius of the sphere
Now using this formula I figured Q would equal $\frac{4piR^3rho}{3}$

so I used this in the numerator for Q and also plugged in r= R/2 and you get

$\frac{AR^3}{6epsilon}$

which is *close* to the right answer, but not quite... where does this standard formula fall apart in reasoning for this particular situation?
 flyboy6212010-11-02 21:12:58 I think you are assuming constant charge density, which is not the case here.
hisperati
2007-10-31 20:29:18
To find the charge we need to integrate a charge density, that is per unit volume. The answers correspond to multipliying by a surface area ( I am not talking about the left side of Guass's Law, where it is obvious surface area is correctly used). Why isn't the charge density multiplied by $\frac{4}{3} pi R^3$ ?
 blah222008-03-14 18:01:46 Yea I'm confused too
 blah222008-03-14 18:27:59 Ah I see, for anyone else confused by this in the future, it's just a method of getting the volume of a sphere through integration. Divide the sphere into lots of shells of thickness dr, each with surface area $4 \Pi r^2$. Thanks so much Yosun!
 flyboy6212010-11-02 21:14:34 You can't just multiply the charge density by the enclosed volume because the density is not constant. You have to integrate.
azot
2007-10-23 06:04:01
the answer to this problem is wrong .
E shoud be equal to A*R^3/32*5
 azot2007-10-23 06:05:00 sorry I was wrong :) Everything is ok
 kevinjay152007-10-23 18:24:01 ETS is never wrong.

hmm. I'm just wondering... I had the following formula memorized for similar situations: E=$\frac{Qr}{4piepsilonR^3}$ Where little r is the radius of the point from the center and R is the total radius of the sphere Now using this formula I figured Q would equal $\frac{4piR^3rho}{3}$ so I used this in the numerator for Q and also plugged in r= R/2 and you get $\frac{AR^3}{6epsilon}$ which is *close* to the right answer, but not quite... where does this standard formula fall apart in reasoning for this particular situation?
LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$